Let f(x) = `{{:( x^(3) + x^(2) - 10 x ,, -1 le x lt 0) , (sin x ,, 0 le x lt x//2) , (1 + cos x ,, pi //2 le x le x ):}` then f(x) has

Options

A. locla maximum at `x = pi/2`
B. local minimum at `x =pi/2`
C. absolute maximum at `x =0`
D. absolute maximum at `x=-1`

Answer

To solve the problem, we need to analyze the piecewise function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} x^3 + x^2 - 10x & \text{for } -1 \leq x < 0 \\ \sin x & \text{for } 0 \leq x < \frac{\pi}{2} \\ 1 + \cos x & \text{for } \frac{\pi}{2} \leq x \leq x \end{cases} \]

We will analyze each piece of the function to determine the nature of \( f(x) \) in terms of local maxima, local minima, and absolute maxima.

### Step 1: Analyze \( f(x) \) for \( -1 \leq x < 0 \)

1. **Function**: \( f(x) = x^3 + x^2 - 10x \)
2. **Find the derivative**:
\[ f'(x) = 3x^2 + 2x - 10 \]
3. **Find critical points by setting the derivative to zero**:
\[ 3x^2 + 2x - 10 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 3 \cdot (-10)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 120}}{6} = \frac{-2 \pm \sqrt{124}}{6} = \frac{-2 \pm 2\sqrt{31}}{6} = \frac{-1 \pm \sqrt{31}}{3} \]
This gives us two critical points:
\[ x_1 = \frac{-1 - \sqrt{31}}{3}, \quad x_2 = \frac{-1 + \sqrt{31}}{3} \]

4. **Evaluate the sign of \( f'(x) \)**:
- For \( x < x_1 \), \( f'(x) > 0 \) (increasing).
- Between \( x_1 \) and \( x_2 \), \( f'(x) < 0 \) (decreasing).
- For \( x > x_2 \), \( f'(x) > 0 \) (increasing).

5. **Behavior in the interval**:
- Since \( x_1 < -1 \) and \( x_2 > 0 \), \( f(x) \) is decreasing on \( [-1, 0) \).
- Thus, the absolute maximum occurs at \( x = -1 \).

### Step 2: Analyze \( f(x) \) for \( 0 \leq x < \frac{\pi}{2} \)

1. **Function**: \( f(x) = \sin x \)
2. **Find the derivative**:
\[ f'(x) = \cos x \]
3. **Critical points**:
- \( f'(x) = 0 \) when \( \cos x = 0 \) which occurs at \( x = \frac{\pi}{2} \).
4. **Behavior**:
- \( f'(x) > 0 \) for \( 0 < x < \frac{\pi}{2} \) (increasing).
- Thus, the maximum at \( x = \frac{\pi}{2} \) is \( f\left(\frac{\pi}{2}\right) = 1 \).

### Step 3: Analyze \( f(x) \) for \( \frac{\pi}{2} \leq x \)

1. **Function**: \( f(x) = 1 + \cos x \)
2. **Find the derivative**:
\[ f'(x) = -\sin x \]
3. **Behavior**:
- \( f'(x) < 0 \) for \( x \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \) (decreasing).
- Thus, \( f(x) \) is decreasing after \( x = \frac{\pi}{2} \).

### Conclusion

- **Local maxima**: At \( x = \frac{\pi}{2} \) (since \( f'(x) \) changes from positive to negative).
- **Absolute maxima**: At \( x = -1 \) (the highest value of \( f(x) \) in the interval).
- **Local minima**: There are no local minima in the given intervals.

### Final Answer

- \( f(x) \) has a local maximum at \( x = \frac{\pi}{2} \) and an absolute maximum at \( x = -1 \).